Problem: A particle moving in the $xy$ -plane has velocity vector given by $v(t)=\left(\cos(2t),\sin(2t)\right)$ for time $t\geq 0$. What is the magnitude of the displacement of the particle between time $t=0$ and $t=\dfrac{\pi}{4}$ ? Round to the nearest tenth.
To find the magnitude of the displacement of the particle, we should first find the particle's horizontal displacement $\Delta x$ and the particle's vertical displacement $\Delta y$. Then we can find the magnitude of the displacement using the distance formula: $\text{Magnitude of displacement }=\sqrt{(\Delta x)^2+(\Delta y)^2}$ The particle's horizontal displacement can be found by taking the definite integral of the horizontal component of $v(t)$ between time $t=0$ and $t=\dfrac{\pi}{4}$ : $\Delta x=\int_{0}^{\tfrac{\pi}{4}} \cos(2t)\,dt=\dfrac{1}{2}$ The particle's vertical displacement can be found by taking the definite integral of the vertical component of $v(t)$ between time $t=0$ and $t=\dfrac{\pi}{4}$ : $\Delta y=\int_{0}^{\tfrac{\pi}{4}} \sin(2t)\,dt=\dfrac{1}{2}$ Now we can find the magnitude of the displacement: $\begin{aligned} &\phantom{=}\sqrt{(\Delta x)^2+(\Delta y)^2} \\\\ &=\sqrt{\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^2} \\\\ &\approx 0.7 \end{aligned}$ In conclusion, the magnitude of the displacement of the particle between time $t=0$ and $t=\dfrac{\pi}{4}$ is $0.7$ units.